Quantum confinement - Optical Properties of Solids 2nd Ed by Mark Fox

(6.1) Substitute into eqn 6.3 with ∆x = 10⁻⁶m and m = 0.1m0 to obtain T . 0.01 K.

(6.2) The energy difference between the n = 1 and n = 2 levels is worked out from eqn 6.13 to be 3~²π²/2m^∗d². On setting this energy difference to be equal to kBT /2, we then derive the required result, namely

d = s

3~²π² m^∗kBT .

On substituting into this formula with T = 300 K, we find d = 9.3 nm for m^∗= m0, and d = 30 nm for m^∗= 0.1m0.

On comparing with eqn 6.4, it is apparent that d =√

3π²∆x. This shows that the two criteria used to determine when quantum size effects are important give the same answer, apart from a numerical factor of ∼ 5.

The differing numerical factor is to be expected, given the approximate nature of the criteria.

(2p/L)² k

Figure 20: Grid of allowed k states in a two-dimensional material of area L², as required for the solution of Exercise 6.3.

(6.3) The x and y components of the k vector must each satisfy the criterion exp(ikL) = 1, which implies k = integer × 2π/L. The possible values of the k vector therefore form a regular grid in k-space as shown in Fig. 20, with a grid-spacing of 2π/L. The area per k-state is (2π/L)², and the

density of states for an area L² is therefore L²/(2π)². This implies that the density of states in k-space for a unit area of crystal is 1/(2π)². It is also apparent from Fig. 20 that the area of k-space enclosed by the increment k → k + dk is 2πkdk. If we call the number of k-states enclosed by this area g^L(k)dk, we then find:

g^L(k)dk = 2πkdk (2π/L)² = L²

2πkdk . Hence for a unit area of crystal we have:

g2D(k)dk = k 2πdk .

The density of states in energy space can be found from eqns 3.13–14. (The factor or two accounts for the fact that spin 1/2 particles have two spin states for each k-state.) With E(k) = ~²k²/2m, we have dE/dk = ~²k/m, and hence

g2D(E) = 2g2D(k)

dE/dk = 2k/2π

~²k/m= m π~², as required.

(6.4) In one dimension the k states must all lie on a single axis (say the x axis). Let L be the length of the wire. The Born–von Karman boundary conditions require that exp(ikL) = 1, and hence that k = integer × 2π/L.

The density of states in k space for a sample of unit length is therefore 1/2π. We thus have:

g1D(k)dk = 1 2πdk .

The density of states in energy space is then found from eqns 3.13–14 with E = ~²k²/2m. This gives:

g1D(E) = 2g1D(k)

dE/dk = 2/2π

~²k/m = m

π~²k =³ m 2~²π²

´_1/2 E^−1/2,

as required.

(6.5) The depth of the potential well enters eqn 6.26 via ξ, which is defined in eqn 6.27. The function on the right hand side of eqn 6.26 decreases from (m^∗_w/m^∗_b)^1/2√

ξ at x = 0 to zero at x =√

ξ. (See, for example, Fig. 6.5, for the case with ξ = 13.2.) The function on the RHS of eqn 6.26 will therefore always cross the x tan x function between 0 and π/2, no matter how small ξ is.

(6.6) We follow the method of Example 6.1. We have µm^∗_w

m^∗_b

¶_1/2

= µ0.34

0.5

¶_1/2

= 0.82 , and

ξ = 0.34m0× (10⁻⁸)²× 0.15 eV

2~² = 33.5 .

0 2 4 6 0

2 4 6

x y= x tan x

y= 0.82 (33.5-x²)^1/2

x=1.30

0 2 4 6

x y= x tan x

y= 0.82 (33.5-x²)^1/2

x=1.30

Figure 21: Graphical solution required for Exercise 6.6.

Hence we must solve:

x tan x = 0.82p

33.5 − x².

The two functions are plotted in Fig. 21, which shows that the solution is x = 1.30. The energy is then found from

E =2~²x²

m^∗_wd² = 7.6 meV ,

for d = 10 nm. The equivalent energy for an infinite well is calculated from eqn 6.13 to be 11 meV.

(6.7) (a) The wave functions of an infinite potential well form a complete or-thonormal basis, with

Z _+∞

−∞

ϕ^∗_nϕn⁰dz = δnn⁰,

where δnn⁰ is the Kronecker delta function:

δnn⁰ = 1 if n = n⁰

= 0 if n 6= n⁰.

It is apparent from eqn 6.11 that the wave functions depend only on n and are therefore identical for electrons and holes with the same n. Hence the orthonormality condition applies irrespective of whether ϕn and ϕn⁰

are electron or hole wave functions, or a mixture. Hence:

Mnn⁰ = 1 if n = n⁰

= 0 if n 6= n⁰.

This result can also be derived by explicit (and somewhat tedious) substi-tution of the wave functions from eqn 6.11 into the formula for Mnn⁰. (b) This result follows from parity arguments. The wave functions of a finite well have well-defined parities as a consequence of the inversion

symmetry about the centre of the well. Wave functions with odd n have even parity, while those with even n have odd parity. Hence the product ϕ^∗_enϕ^∗_hn0 has even parity if n − n⁰is even and has odd parity for odd n − n⁰. The integral of an odd function from −∞ → +∞ is zero, and so the overlap integral is zero if ∆n is equal to an odd number.

n Electron Heavy hole Light hole

1 225 30 188

2 898 120 750

Table 1: Confinement energies in meV calculated for an infinite quantum well of width 5 nm, as required for Exercise 6.8.

(6.8) The confinement energies of the electrons, heavy holes and light holes calculated from eqn 6.13 are given in Table 1. With an infinite well, we only need consider ∆n = 0 transitions. The threshold photon energies for these transitions are given by eqn 6.39, or the equivalent for higher values of n or for the light holes. Two transitions fall in the range 1.4 → 2.0 eV:

• Heavy hole 1 → electron 1 at 1.679 eV,

• Light hole 1 → electron 1 at 1.837 eV.

For each transition we expect a step at the threshold energy as in Fig. 6.9.

In 2-D materials the joint density of states is proportional to the electron-hole reduced mass µ (see eqn 6.41), and hence the relative height of the heavy-hole and light-hole transition steps is in proportion to their reduced masses, that is 0.059 : 0.036. Hence the final spectrum would appear as in Fig. 22.

1.4 1.6 1.8 2.0

0.0 0.1

Absorption(arb.units)

Photon energy (eV) hh1 ® e1

lh1 ® e1 1.679 eV

1.837 eV

1.4 1.6 1.8 2.0

0.0 0.1

Absorption(arb.units)

Photon energy (eV) hh1 ® e1

lh1 ® e1 1.679 eV

1.837 eV

Figure 22: Absorption spectrum for Exercise 6.8

(6.9) (a) In finite wells the confinement energies are reduced compared to those of an infinite well of the same width. Hence the transition energies would be lower. Furthermore, transitions that are forbidden in infinite wells become weakly allowed, such as the hh3 → e1 transition. This transition would fall within the observed energy range, as might also some n = 2 transitions.

(b) Peaks would appear below the steps in the absorption spectrum due to excitonic absorption. The difference in energy between the peak and the continuum absorption would be equal to the exciton binding energy.

(6.10) (a) To prove normalization, we must show thatRR

ψ^∗ψ dA = 1. For the given wave function we have:

Z _∞ We can then calculate hEivar from:

hEivar =

(c) On differentiating hEivarwith respect to ξ, we find that hEivarachieves its minimum value of Emin= −µe⁴/8(π²0²r~)²for ξ = 2π~²²0²r/µe². The minimum energy is four times larger than the bulk exciton binding energy found in Exercise 4.4.

(d) In part (c) we found ξmin = 2π~²²0²r/µe². This can be written in terms of the bulk exciton radius aX defined in eqn 4.2 as:

ξmin= aX/2 .

Hence the radius of the exciton in 2-D is half the radius of the bulk.

(6.11) At d = ∞ we have bulk GaAs, while at d = 0 we have bulk AlGaAs.

For intermediate values of d, we have a GaAs quantum well exciton with an enhanced binding energy. In an ideal 2–D system we would expect

four times the binding energy of bulk GaAs (i.e. 16 meV), but in realistic systems, the enhancement might be smaller due to the imperfect quantum confinement of the finite-height AlGaAs barriers. Thus as d is reduced from ∞, the binding energy increases from 4 meV, going through a peak, and then dropping to 6 meV as d → 0. The height of the peak might typically be around 10 meV.

1.58 1.60 1.62 1.64 0.0

1.0

Energy (eV)

PLEintensity

hh1 ® e1 exciton

lh1 ® e1 exciton hh1 ® e1 band edge

lh1 ® e1 band edge

1.58 1.60 1.62 1.64 0.0

1.0

Energy (eV)

PLEintensity

hh1 ® e1 exciton

lh1 ® e1 exciton hh1 ® e1 band edge

lh1 ® e1 band edge

Figure 23: Interpretation of the data in Fig. 6.23 as required for Exercise 6.12.

(6.12) (a) See Section 5.3.5.

(b) The interpretation of the principal features in the data is shown in Fig. 23. For both heavy- and light-hole transitions, we expect to observe a peak due to excitonic absorption followed by a step at the band edge. The two strong peaks observed in the data correspond to the excitons for the hh1 → e1 and lh1 → e1 transitions, while the flat absorption bands above both excitons correspond to the interband transitions. These interband absorption bands are flat because the density of states is independent of the energy in 2-D systems (see eqn 6.41.)

(c) The hh1 → e1 interband continuum starts at 1.592 eV. In the infinite well model, the transition energy is given by eqn 6.42 with n = 1, which, on using Eg = 1.519 eV, m^∗_e = 0.067m0 and m^∗_hh = 0.5m0, implies d = 9.3 nm. The real well width would be smaller, because the infinite well model overestimates the confinement energy.

(d) The binding energies can be read from Fig. 6.23 as the energy gap between the exciton peak and the appropriate band edge. With the tran-sitions identified as in Fig. 23, we find binding energies of 11 meV for the heavy holes and 12 meV for light holes. For a perfect 2-D system we would expect 4 × R^Xfor GaAs, i.e. 16.8 meV. The experimental values are lower because a real quantum well is not a perfect 2-D system.

(6.13) By considering Fig. 6.12 we would expect three different regimes to apply.

1. For photon energies in the range set by eqn 6.43, that is:

Eg+ Ee1+ Ehh1≤ ~ω < Eg+ Ee1+ Elh1,

we can only excite hh1 → e1 transitions. These only create MJ =

−1/2 spin down electrons, and so we expect Πe= −100%.

2. For photon energies in the range:

Eg+ Ee1+ Elh1≤ ~ω < Eg+ Ee1+ Eso1,

where Eso1is the confinement energy of the first split-off hole band, we can excite both hh1 → e1 and lh1 → e1 transitions, which create oppositely polarized electrons. Hence we expect the magnitude of the electron spin polarization to decrease. If the matrix elements are the same as in the bulk, the heavy-hole transitions will be three times stronger than the light-hole transitions, and this gives an electron spin polarization of −50%. (See Section 3.3.7.)

3. For photon energies in the range:

~ω > Eg+ Ee1+ Eso1,

we expect that no spin polarization will be be created. The extra contribution of the split-off hole band exactly cancels the net polar-ization created by the stronger weight of the heavy-hole transition.

This can be understood in general terms by realizing that the elec-tron spin polarization is created by spin–orbit coupling, so that when the excess photon energy exceeds the spin–orbit interaction energy

∆, then the effect will be negligible.

Experimental data showing these effects may be found in Vi˜na, J. Phys.:

Condens. Matter 11, 5929 (1999), Fig. 2. The basic trends discussed above are reproduced in the experimental data, although there are some important differences. This is because we really should work in the ex-citonic rather than single-particle picture, and also because the relative weight of the heavy- and light-hole transitions is not exactly 3:1. See the discussion in Vi˜na’s article for further details.

(6.14) (a) With H⁰ = +ezEz, we have:

∆E⁽¹⁾= eEz

Z _+∞

−∞

ϕ^∗zϕ dz .

Since z is an odd function, and ϕ^∗ϕ ≡ |ϕ|² is an even one, the integral is zero.

(b) With H⁰= +ezEz, the second-order energy shift is given by:

∆E⁽²⁾= e²E²_z X∞ n=2

|h1|z|ni|² E1− En

Since the wave functions have parity (−1)ⁿ⁺¹, and z is an odd parity operator, all the terms with odd n are zero. Hence we have:

∆E⁽²⁾ = e²E²_z|h1|z|2i|² E1− E2

+ e²E²_z|h1|z|4i|² E1− E4

+ · · · .

Now the terms with n ≥ 4 are much smaller than the term with n = 2.

(This can be verified by working through the integrals, but it is fairly

obvious given the larger denominator.) Hence we only need to consider the first term:

∆E⁽²⁾= e²E²_z|h1|z|2i|² E1− E2 .

On substituting the energies for an infinite well from eqn 6.13, this be-comes:

∆E⁽²⁾ = −2e²E²_zm^∗d²

3~²π² |h1|z|2i|².

Now, on redefining the origin so that the quantum well runs from z = 0 to z = +d, we have:

h1|z|2i =

Z _+∞

−∞

ϕ^∗₁z ϕ2dz ,

= 2

d Z _d

sin(πz/d) z sin(2πz/d) dz ,

= −16d 9π². Hence we find

∆E⁽²⁾ = −2e²E²_zm^∗d² 3~²π² ×

¯¯

¯¯−16d 9π²

¯¯

= −24 µ 2

3π

¶₆

e²E²_zm^∗d⁴

~² , as required.

(6.15) (a) Figure 6.13 predicts that the energy of the hh1→e1 transition will shift from 1462.4 meV to 1438.5 meV in a 10 nm quantum well on in-creasing the field from zero to 1 × 10⁷ V/m. This corresponds to a red shift of 14.1 nm. This can be compared to the experimental red shift of 10.5 nm for a 9 nm well for a voltage change of 10 V. Excitonic effects are not included in the calculation of Fig. 6.13, but this is not expected to make a large difference. Two main factors account for the difference between the experimental and theoretical results.

• The well width for the experiment was smaller by a factor of 0.9.

From eqn 6.47 we would then expect the Stark shift to be smaller by a factor (0.9)⁴= 0.66.

• For the experimental data, we can calculate the field from eqn 6.52.

From this we find that Ez changes from 0.15 × 10⁷ V/m to 1.15 × 10⁷ V/m, which should be compared to the theoretical data, which corresponds to changing Ezfrom 0 → 1 × 10⁷ V/m. The field change is the same in both cases, but because the Stark shift is quadratic at low fields, we do not expect the Stark shifts to be the same. If the Stark shift remained quadratic at all field strengths, we would expect the experiment shift to be larger by a factor (1.15²−0.15²)/(1²−0²) = 1.30.

Putting these two factors together, we expect the experimental Stark shift to be smaller than the theoretical one by a factor of 0.9⁴× 1.3 = 0.85. The actual ratio is (10.5/14.1) = 0.74. This difference is easily explained by

the fact that the Stark shift does not remain quadratic at all fields, and so the factor due to the different fields should be smaller.

(b) We assume that the Stark shift is quadratic, and hence that ∆E ∝ E²_z. We work out the field strength from eqn 6.52, which implies Ez = 11.5 MV/m at 10 V and Ez = 6.5 MV/m at 5 V. For small shifts we have

∆λ ∝ ∆E, and so we find:

∆λ(5 V) = µ 6.5

11.5

¶2

∆λ(10 V) = 0.32 × 10.5 nm = 3.4 nm .

(c) The wavelength red shift of 10.5 nm corresponds to an energy shift of –18 meV. This energy shift is related to the net electron-hole displacement h∆zi by:

∆E = −pzEz= −eh∆ziEz. With Ez= 11.5 MV/m, we thus obtain h∆zi = 1.6 nm.

Sample A Sample B

Ez (MV/m) ∆Ecalc ∆Eexp ∆Ecalc ∆Eexp

3 1.5 1 15 6

6 5.9 5 62 27

9 13 13 139 54

Table 2: Comparison of the calculated and measured Stark shifts (in meV) for the two samples discussed in Exercise 6.16.

(6.16) The experimental data is taken from Polland et al., Phys. Rev. Lett. 55, 2610 (1985). We analyse it by using the energy shift calculated by second-order perturbation theory in Exercise 6.14(b). Since we are considering an electron-hole transition, we must add together the Stark shifts of the electrons and holes, giving:

∆E⁽²⁾= −24 µ 2

3π

¶₆ e²E²_zd⁴

~² (m^∗_e+ m^∗_hh) .

The calculated shifts using m^∗_e = 0.067m0and m^∗_hh= 0.5m0are compared to the experimental ones in Table 2. It is apparent that the model works well for sample A, but not for sample B. The model breaks down when the size of the Stark shift becomes comparable to the energy splitting of the unperturbed hh1 and hh2 levels. This is essentially the same criterion as for the transition from the quadratic to the linear Stark effect in atomic physics. In sample B, we are in this regime at all the fields considered.

(6.17) At Ez = 0 the quantum well is symmetric about the centre of the well.

The electron and hole states therefore have a definite parity with respect to inversion about z = 0. The parity is (−1)⁽ⁿ⁺¹⁾, and the electron–hole overlap given by eqn 6.36 is zero if ∆n is odd: see Exercise 6.7(b). At finite Ez, the inversion symmetry is broken, and the states no longer have a definite parity. Therefore, the selection rule based on parity no longer holds.

(6.18) With the confinement energy E ∝ d⁻², we have dE/dd ∝ −2/d³, and hence we expect:

∆E

E = −2∆d d .

A ±5% change in d is thus expected to give ∆E/E = ±10%. With E = 0.1 eV, we then expect a full-width broadening of 0.02 eV. This is comparable to the linewidth observed in the 10 K data shown in Fig. 6.16.

A ±5% variation in d corresponds more or less to a fluctuation of one atomic layer. Such “monolayer” fluctuations are unavoidable in the crystal growth. The linewidth at room temperature is further broadened by the thermal spread of the carriers in the bands.

(6.19) We assume infinite barriers and use eqn 6.53 to calculate the emission energy. With ~ω = 0.80 eV, Eg = 0.75 eV, and µ = 0.038m0, we find d = 14 nm. In reality, the quantum well would have to be narrower to compensate for the imperfect confinement of the barriers.

(6.20) (a) z is an odd function with respect to inversion about z = 0. The integral from −∞ to +∞ will therefore be zero unless ϕ^∗_nϕn⁰ is also an odd function, which requires that the wave functions must have different parities. Since the wave functions have parities of (−1)ⁿ⁺¹, the condition is satisfied if n is an even number and n⁰ odd, and vice versa. Hence ∆n must be equal to an odd number.

(b) The strength of the intersubband transitions is proportional to the square of the matrix elements. These matrix elements can be evaluated by substituting the wave functions from eqn 6.11. On redefining the origin so that the quantum well runs from z = 0 to z = +d, we find:

h1|z|2i = 2 d

Z _d

sin(πz/d) z sin(2πz/d) dz = − 16 9π²d , and

h1|z|4i = 2 d

Z _d

sin(πz/d) z sin(4πz/d) dz = − 4 45π²d .

Hence the 1 → 4 transition is weaker than the 1 → 2 transition by a factor [(4/45)/(16/9)]²= [1/20]²= 2.5 × 10⁻³.

It is apparent from Fig. 6.17 that the wavelength of the 1 → 2 transition is given by

λ = E2− E1= 3π²~² 2m^∗_ed²,

where we used the infinite well energies of eqn 6.13 in the second equality.

On inserting m^∗_e = 0.067m0and d = 20 nm, we find hc/λ = 0.042 eV, and hence λ = 29 µm.

(6.21) Consider a ray incident at angle θ to the normal as shown in Fig. 24.

The ray will be refracted according to Snell’s law, with sin θ

sin θ⁰ = n ,

q¢ q

semiconductor, refractive index n air

n= 1 E

E¢

quantum well

q¢ q

semiconductor, refractive index n air

n= 1 E

E¢

quantum well

Figure 24: Refraction of light entering a semiconductor containing a quantum well, as discussed in Exercise 6.21.

where θ⁰ is the angle inside the crystal. For intersubband transitions, we are interested in the z component of the electric field of the light at the quantum well, namely:

Ez= E⁰sin θ⁰= E⁰sin θ/n .

If I0 is the incident intensity, and there are no intensity losses at the surface, then the intensity in the z component at the quantum well is given by

Iz= I0(sin θ/n)²,

since the intensity is proportional to E². Hence the fraction of the power of the beam in the z polarization at the quantum well is (sin θ/n)². This fraction has a maximum value of 1/n² for θ = 90^◦. Therefore even if we completely absorb all the z polarized light by intersubband transitions, and we use glancing incidence, we can only remove a fraction of 1/n² of the power in the incident beam. This fractional absorption is equal to 9%

if n = 3.3.

nx ny nx (n²_x+ n²_y+ n²_z) g

1 1 1 3 1

2 1 1 6 3

2 2 1 9 3

3 1 1 11 3

2 2 2 12 1

3 2 1 14 6

3 2 2 17 3

Table 3: Quantum numbers of the energy levels for a cubic quantum dots in or-der of increasing energy, as discussed in Exercise 6.22. g denotes the degeneracy excluding spin.

(6.22) For a cubic dot, the energies of the quantized levels are given by eqn 6.54 with dx= dy= dz = d, implying:

E(nx, ny, nz) = π²~²

2m^∗d²(n²_x+ n²_y+ n²_z) ,

where nx, ny and nz are integers with a minimum value of 1. As demon-strated by Table 3, the quantized levels occur at energies of 3, 6, 9, 11, 12, 14, 17,. . . in units of h²/8m^∗d².

The degeneracy factor g is worked out by finding the number of per-mutations of (nx, ny, nz) that can give the same energy. The ground state is unique, but for the first excited we can have (2, 1, 1), (1, 2, 1) and (1, 1, 2), and similarly for the levels at 9, 11, and 17 h²/8m^∗d². The level at 14 h²/8m^∗d² has a degeneracy of 6 because there are 3! = 6 ways of arranging the numbers 3, 2 and 1.

(6.23) The radial equation for a spherical dot is given by eqn 6.57. If the con-fining potential is V0, then we can put V (r) = −V0for r ≤ R0. Therefore, when the particle is inside the dot (i.e. r ≤ R0), the radial wave function for states with l = 0 must satisfy:

Hence if R(r) = sin kr/r, then substitution into the radial equation gives:

~²k² where n is an integer. Hence:

E = −V0+~²n²π² 2m^∗R²₀.

The confinement energy relative to the bottom of the potential well at

−V0 is thus:

En= ~²n²π² 2m^∗R²₀.

(6.24) For the dots to have the same volume, we require that:

d³=4 3πR³₀,

where d is the cube size and R0 is the radius of the spherical dot. This implies that (R0/d) = (3/4π)^1/3. From Exercise 6.22 we find that the confinement energy of the first level in a cubic dot with infinite barriers is E₁^cubic = 3~²π²/2m^∗d². For a spherical dot eqn 6.58 gives E₁^spherical=

The reason why the cube has the higher energy can be understood by reference to Fig. 25. In the infinite well approximation, the wave functions are proportional to sine waves with nodes at the edges for both the cube and the sphere. It is apparent from Fig. 25 that if the cube and sphere have the same volume, then the wavelength of the sine wave in the cube

In document Optical Properties of Solids 2nd Ed by Mark Fox (Page 57-73)